1140. Stone Game II.cpp

// 1140. Stone Game II
// Solved
// Medium
// Topics
// Companies
// Hint
// Alice and Bob continue their games with piles of stones.  There are a number of piles arranged in a row, and each pile has a positive integer number of stones piles[i].  The objective of the game is to end with the most stones. 

// Alice and Bob take turns, with Alice starting first.  Initially, M = 1.

// On each player's turn, that player can take all the stones in the first X remaining piles, where 1 <= X <= 2M.  Then, we set M = max(M, X).

// The game continues until all the stones have been taken.

// Assuming Alice and Bob play optimally, return the maximum number of stones Alice can get.

 

// Example 1:

// Input: piles = [2,7,9,4,4]
// Output: 10
// Explanation:  If Alice takes one pile at the beginning, Bob takes two piles, then Alice takes 2 piles again. Alice can get 2 + 4 + 4 = 10 piles in total. If Alice takes two piles at the beginning, then Bob can take all three piles left. In this case, Alice get 2 + 7 = 9 piles in total. So we return 10 since it's larger. 
// Example 2:

// Input: piles = [1,2,3,4,5,100]
// Output: 104
 

// Constraints:

// 1 <= piles.length <= 100
// 1 <= piles[i] <= 104

class Solution {
public:
    int stoneGameII(vector<int>& piles) {
        vector<vector<int>> memo(piles.size(), vector<int>(piles.size()));
        vector<int> suffixSum = piles;
        for (int i = suffixSum.size() - 2; i >= 0; --i)
            suffixSum[i] += suffixSum[i + 1];
        return maxStones(suffixSum, 1, 0, memo);
    }

    int maxStones(vector<int>& suffixSum, int maxTillNow, int currIndex,
                  vector<vector<int>>& memo) {
        if (currIndex + 2 * maxTillNow >= suffixSum.size())
            return suffixSum[currIndex];
        if (memo[currIndex][maxTillNow] > 0) return memo[currIndex][maxTillNow];
        int res = INT_MAX;

        for (int i = 1; i <= 2 * maxTillNow; ++i) {
            res = min(res, maxStones(suffixSum, max(i, maxTillNow),
                                     currIndex + i, memo));
        }
        memo[currIndex][maxTillNow] = suffixSum[currIndex] - res;
        return memo[currIndex][maxTillNow];
    }
};